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Juliette [100K]
3 years ago
15

What is the total number of calories of heat energy absorbed when 10 grams of water is vaporized at its normal boiling point

Chemistry
1 answer:
Darya [45]3 years ago
6 0

<u>Answer:</u> The amount of energy absorbed by water is 5390 Calories

<u>Explanation:</u>

To calculate the amount of heat absorbed at normal boiling point, we use the equation:

q=m\times L_{vap}

where,

q = amount of heat absorbed = ?

m = mass of water = 10 grams

L_{vap} = latent heat of vaporization = 539 Cal/g

Putting values in above equation, we get:

q=10g\times 539Cal/g=5390Cal

Hence, the amount of energy absorbed by water is 5390 Calories

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How many grams of CaCl2 are in 250 mL of 2.0 M CaCl2?
Tom [10]
The answer is:  " 56 g CaCl₂ " .
__________________________________________________________

Explanation:
__________________________________________________________
2.0 M CaCl₂  = 2.0 mol CaCl₂ / L  ; 

Since: "M" = "Molarity" (measurement of concentration); 

                  = moles of solute per L {"Liter"} of solution.
__________________________________________________________
Note the exact conversion:  1000 mL = 1 L . 

Given: 250 mL ;   

250 mL = ?  L  ?  ;  


250 mL * (1 L / 1000 L) =  (250/1000) L = 0.25 L . 
___________________________________________________________
 
(2.0 mol CaCl₂ / L ) * (0.25L) = (2.0) * (0.25) mol  = 0.50 mol CaCl₂ ;

We have: 0.50 mol CaCl₂ ;  Convert to "g" (grams):

→ 0.50 mol CaCl₂  .
___________________________________________________________
1 mol CaCl₂ = ? g ?

From the Periodic Table of Elements:

1 mol Ca = 40.08 g

1 mol Cl  =  <span>35.45 g .
</span>
There are 2 atoms of Cl in " CaCl₂ " ;  

→ Note the subscript, "2", in the " Cl₂ " ; 
__________________________________________________________
So, to calculate the molar mass of "CaCl₂" :

40.08 g  +  2(35.45 g) = 

40.08 g  +  70.90 g = 110.98 g ;  round to 4 significant figures; 

                                 → round to 111 g/mol .
__________________________________________________________
So:

→  0.50 mol CaCl₂  = ? g CaCl₂  ? ; 

→  0.50 mol CaCl₂ * (111 g CaCl₂ / mol CaCl₂) ;

                                             = (0.50) * (111 g) CaCl₂ ;

                                             =  55.5 g CaCl₂  ;

                                                → round to 2 significant figures; 

                                                →  56 g CaCl₂ .
___________________________________________________________
The answer is:  " 56 g CaCl₂ " .
___________________________________________________________
6 0
3 years ago
Read 2 more answers
How many electrons does hydrogen need to satisfy the octet rule?
quester [9]
Just 2 valence electrons.

Hydrogen already has one to start with, as well. With the exception of hydrogen and helium, all other atoms need 8 valence e-
4 0
3 years ago
What products are always made when any carbon-based fuel
Elodia [21]
Answer:
d) carbon dioxide and water
8 0
2 years ago
How many moles are in 50 g of CO2
murzikaleks [220]

Answer:

1.1 mol

Explanation:

n=m/M, where n is moles, m is mass, and M is molar mass.

M of CO2 = 12.01+16.00+16.00 = 44.01g/mol

n=50g/44.01g/mol

n = 1.13610543 mol

n ≈ 1.1 mol

Hope that helps

8 0
2 years ago
A 0.223 mole sample of gas is held at 33.0 C and 2.00 atm, What's the volume of the gas? R = 0.0821 L atm / mol K answer soon il
Ghella [55]

Answer:

The volume of the gas is 2.80 L.

Explanation:

An ideal gas is a theoretical gas that is considered to be made up of point particles that move randomly and do not interact with each other. Gases in general are ideal when they are at high temperatures and low pressures.

The Pressure (P) of a gas on the walls of the container that contains it, the Volume (V) it occupies, the Temperature (T) at which it is located and the amount of substance it contains (number of moles, n) are related from the equation known as Equation of State of Ideal Gases:

P*V = n*R*T

where R is the constant of ideal gases.

In this case:

  • P= 2 atm
  • V= ?
  • n=0.223 moles
  • R= 0.0821 \frac{L*atm}{mol*K}
  • T=33 °C= 306 °K (being O°C= 273°K)

Replacing:

2 atm* V= 0.223 moles*0.0821 \frac{L*atm}{mol*K}* 306 K

Solving:

V=\frac{0.223 moles*0.0821\frac{L*atm}{mol*K} * 306 K}{2 atm} \\

V= 2.80 L

<u><em>The volume of the gas is 2.80 L.</em></u>

7 0
3 years ago
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